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Question 13.6: Determining the Equilibrium Constant Kp Methane (CH4) reacts......

Determining the Equilibrium Constant K_p

Methane (CH_4) reacts with hydrogen sulfide to yield H_2 and carbon disulfide, a solvent used in manufacturing rayon and cellophane:

\mathrm{CH}_4(g)+2 \mathrm{H}_2 \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_2(g)+4 \mathrm{H}_2(g)

What is the value of K_p at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.20 atm of CH_4, 0.25 atm of H_2S, 0.52 atm of CS_2, and 0.10 atm of H_2?

STRATEGY

Write the equilibrium equation by setting K_p equal to the equilibrium-constant expression using partial pressures. Put the partial pressures of products in the numerator and the partial pressures of reactants in the denominator, with the pressure of each substance raised to the power of its coefficient in the balanced chemical equation. Then substitute the partial pressures into the equilibrium equation and solve for K_p.

worked example 13.6
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K_{\mathrm{p}}=\frac{\left(P_{\mathrm{CS}_2}\right)\left(P_{\mathrm{H}_2}\right)^{4 \ \leftarrow Coefficient \ of \ H_2}}{\left(P_{\mathrm{CH}_4}\right)\left(P_{\mathrm{H}_2 \mathrm{~S}}\right)^{2 \ \leftarrow \ Coefficient of H_2S}} \\ K_{\mathrm{p}}=\frac{\left(P_{\mathrm{CS}_2}\right)\left(P_{\mathrm{H}_2}\right)^4}{\left(P_{\mathrm{CH}_4}\right)\left(P_{\mathrm{H}_2 \mathrm{~S}}\right)^2}=\frac{(0.52)(0.10)^4}{(0.20)(0.25)^2}=4.2 \times 10^{-3}

Note that the partial pressures must be in units of atmospheres (not mm Hg) because the standard-state partial pressure for gases is 1 atm

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